Power - AP Physics 1

Two students pull identical carts at constant speed $v$ on level ground. Student 1 pulls with force $F$ ; Student 2 pulls with $2F$ . Which statement about their powers is correct?
Answer: Student 2 has greater power because $P=Fv$ and $2F$ at the same $v$ doubles $P$ . This question assesses the concept of power in AP Physics 1, which is the rate at which work is done or energy is transferred. Power is defined as P = W/Δt, but for constant force and velocity, it simplifies to P = Fv, where F is the force and v is the velocity. In this scenario, both students pull carts at the same constant speed v, but Student 2 applies twice the force, 2F, to overcome presumably greater friction or other resistance. Therefore, Student 2's power is 2Fv, which is double that of Student 1's Fv. A common distractor is choice A, which incorrectly assumes power depends only on speed, ignoring the role of force in the P = Fv formula. To approach similar problems, always identify whether power is calculated using energy transfer over time or force times velocity, depending on the given constants.
A constant horizontal force $F$ accelerates a cart from rest on a frictionless track. At time $t$ , the cart's speed is $v$ . At time $2t$ , its speed is $2v$ . How does instantaneous power compare at $t$ and $2t$ ?
Answer: Power at $2t$ is twice that at $t$ because $P=Fv$ . This question assesses the concept of power in AP Physics 1, which is the rate at which work is done or energy is transferred. Instantaneous power for constant force is P = F v, where v changes due to acceleration. With constant acceleration a = F/m, speed at time t is v = a t, and at 2t is 2v, so power at t is F v and at 2t is F (2v) = 2 F v, twice as much. This reflects the increasing rate of kinetic energy transfer as speed grows. Choice C is a distractor, confusing power with kinetic energy, which quadruples but isn't directly the ratio for power. For accelerating objects, calculate instantaneous power using current velocity in P = F v as a reliable method.
A cyclist exerts a forward force $F$ and rides at constant speed $v$ on level ground. Another cyclist exerts force $F/2$ at constant speed $2v$ . Which cyclist produces greater mechanical power?
Answer: They produce the same power because $Fv$ is the same This question tests understanding of mechanical power in cycling. Power equals force times velocity: P = Fv. The first cyclist exerts force F at speed v, producing power P₁ = Fv. The second cyclist exerts force F/2 at speed 2v, producing power P₂ = (F/2)(2v) = Fv = P₁. Both cyclists produce the same mechanical power because the product Fv is identical. Choice B incorrectly assumes greater speed always means greater power without considering the force reduction. To compare power outputs, always calculate the force-velocity product.
A constant horizontal force $F$ moves an object at constant speed $v$ for time $t$ . In a second trial, the force is $2F$ and the speed is $v/2$ for the same time $t$ . Compare average powers.
Answer: They are equal because $P=Fv$ gives the same product This question tests understanding of power using the P = Fv formula. In the first trial, power is P₁ = Fv. In the second trial, with force 2F and speed v/2, power is P₂ = (2F)(v/2) = Fv = P₁. The powers are equal because the product of force and velocity remains constant. Choice A incorrectly assumes larger force always means greater power, ignoring the velocity factor. When force and velocity both change, calculate their product to determine power.
Two machines each transfer energy to a spring. Machine 1 increases the spring's elastic potential energy by $E$ in time $t$ ; Machine 2 increases it by $E$ in time $t/4$ . Which has greater average power?
Answer: Machine 2, because it transfers the same energy in less time. This question assesses the concept of power in AP Physics 1, which is the rate at which work is done or energy is transferred. Average power is P_avg = ΔE/Δt, with both machines increasing elastic potential energy by E. Machine 1 takes time t, so P = E/t; Machine 2 takes t/4, so P = E/(t/4) = 4(E/t), greater power. The shorter time amplifies the rate of energy transfer. Choice C distracts by equating equal energy to equal power, but time differentiates them. For energy transfer comparisons, always divide the energy change by the specific time interval to determine power.
Two machines each do work $W$ on identical objects. Machine A completes the work in time $t$ ; Machine B completes it in time $\tfrac{3}{2}t$ . How do their average powers compare?
Answer: $P_A = \tfrac{3}{2}P_B$ because it takes less time. This question tests understanding of average power calculations. Power is the rate of doing work, defined as P = W/t. For Machine A: $P_A$ = W/t. For Machine B: $P_B$ = W/(3t/2) = W/(1.5t) = (2/3)(W/t) = (2/3) $P_A$ . Therefore, $P_A$ = (3/2) $P_B$ , meaning Machine A has 1.5 times the power of Machine B. Choice B incorrectly inverts the relationship, confusing longer time with greater power. The strategy is to explicitly calculate P = W/t for each case and compare the ratios directly.
Two students lift identical boxes to the same height $h$ . Student 1 lifts in time $t$ ; Student 2 lifts in $2t$ . Which statement about average power is correct?
Answer: Student 1's average power is twice Student 2's. This question tests understanding of power as the rate of doing work. Power is defined as P = W/t, where W is work done and t is time taken. Both students do the same work W = mgh lifting identical boxes to the same height, but Student 1 takes time t while Student 2 takes time 2t. For Student 1: P₁ = W/t, and for Student 2: P₂ = W/(2t) = (1/2)(W/t) = P₁/2. Choice C incorrectly assumes equal power just because work is the same, ignoring that power depends on time. Since P₁ = 2P₂, Student 1's average power is twice Student 2's.
A person carries a box horizontally at constant speed across a level floor. The person's force on the box is vertical and equals $mg$ . What is the mechanical power delivered to the box?
Answer: $0$ This question tests understanding of mechanical power and the direction of force relative to motion. Power is the rate of doing work, calculated as P = F·v·cos(θ), where θ is the angle between force and velocity. The person applies a vertical force (upward) while the box moves horizontally, making θ = 90°. Since cos(90°) = 0, the mechanical power delivered to the box is P = mg·v·0 = 0. Choice A (mgv) incorrectly assumes the force is in the direction of motion. The key insight is that only the component of force in the direction of motion contributes to mechanical power.
Two identical boxes are pulled across a rough floor at constant speeds. Box 1 moves at speed $v$ under pull force $F$ ; Box 2 moves at speed $2v$ under the same pull force $F$ . Which has greater power?
Answer: Box 2, because $P=Fv$ and doubling $v$ doubles power. This question assesses the concept of power in AP Physics 1, which is the rate at which work is done or energy is transferred. For objects moving at constant speed under constant force, power is P = Fv, as work is force times distance and power is work over time. Both boxes are pulled with the same force F, but Box 2 moves at twice the speed, 2v, so its power is F*(2v) = 2Fv, double that of Box 1's Fv. Since the floor is rough but speeds are constant, the force balances friction, and power reflects the rate of energy dissipation. Choice C distracts by claiming equal power due to equal force, ignoring velocity's role in the formula. A useful strategy is to apply P = Fv directly when force and velocity are constant and in the same direction.
Two motors each do the same work $W$ lifting identical loads. Motor 1 takes time $t$ and Motor 2 takes $2t$ . Which motor has greater average power?
Answer: Motor 1, because it does the same work in less time This question tests understanding of power as work done per unit time. Power is defined as P = W/t, where W is work done and t is time taken. Motor 1 does work W in time t, so its power is P₁ = W/t. Motor 2 does the same work W in time 2t, so its power is P₂ = W/(2t) = (1/2)(W/t) = P₁/2. Therefore, Motor 1 has twice the power of Motor 2 because it completes the same work in half the time. Choice B incorrectly assumes equal work means equal power, ignoring the time factor. To compare powers, always consider both work done and time taken: faster completion of the same work means greater power.
A person does $600\ \text{J}$ of work climbing stairs in $10\ \text{s}$ . A friend does $900\ \text{J}$ in $20\ \text{s}$ . Who has greater average power?
Answer: The person, because $600/10 > 900/20$ This question assesses calculating and comparing average power from work and time. Power is the rate of work done, P = W/t. The person does 600 J in 10 s, so P = 600/10 = 60 W, while the friend does 900 J in 20 s, P = 900/20 = 45 W, showing the person has greater power. This demonstrates that higher work does not guarantee higher power if time increases more. Choice B is a distractor focusing only on work magnitude, ignoring the rate aspect. To solve similar problems, compute P explicitly for each case and compare the values directly.
A sled is pulled at constant speed by force $F$ at angle $\theta$ above horizontal. The sled's speed is $v$ . What is the power delivered by the pull?
Answer: $Fv\cos\theta$ This question evaluates power with angled forces at constant speed. Power is the rate of work done, given by P = F · v = F v cosθ, where θ is the angle between force and velocity. Here, the horizontal velocity v and force at θ above horizontal yield P = F v cosθ, as only the horizontal component contributes to work along the direction of motion. This applies because at constant speed, power matches the rate of energy dissipation, like against friction. Choice A, F v, is a distractor that neglects the angle, assuming full force contributes. A strategy is to use the dot product for power when force and velocity are not parallel.
Two identical sleds move at constant speed. Sled A is pulled by force $F$ at speed $v$ ; Sled B by $2F$ at speed $\tfrac{v}{2}$ . How do their powers compare?
Answer: $P_A = P_B$ because $Fv = (2F)\left(\tfrac{v}{2}\right)$ . This question tests understanding of instantaneous power for constant velocity motion. Power is the rate of energy transfer, and for a force applied to an object moving at constant velocity, P = F·v. For Sled A: $P_A$ = F·v. For Sled B: $P_B$ = (2F)·(v/2) = 2F·v/2 = F·v. Therefore, $P_A$ = $P_B$ , showing that doubling force while halving velocity maintains the same power. Choice A incorrectly considers only velocity, ignoring force. The key insight is that power depends on the product of force and velocity, not either factor alone.
A crate is pulled at constant speed $v$ across a rough floor by a horizontal force $F$ for time $t$ . What is the average power delivered by the pulling force?
Answer: $Fv$ This question tests understanding of power as the rate of energy transfer. Power is defined as work done per unit time, or P = W/t. Since the crate moves at constant speed v for time t, the displacement is d = vt, and the work done by force F is W = Fd = F(vt). Therefore, power P = W/t = F(vt)/t = Fv. Choice A (Ft) incorrectly represents work rather than power, missing the division by time. When calculating power for a constant force moving an object at constant velocity, multiply force by velocity: P = Fv.
A crane lifts a mass $m$ straight up at constant speed $v$ . Ignoring air resistance, what is the crane's power output?
Answer: $mgv$ This question examines power output in vertical lifting at constant speed. Power is the rate of work done or energy transfer, and for constant velocity, P = F v. The force required to lift mass m at constant speed v is F = m g to counter gravity, so P = m g v. This applies because the crane transfers energy at a rate matching the gain in gravitational potential energy over time. Choice D, m g h, is a distractor representing work, not power, as it omits the time or velocity factor. A transferable strategy is to identify if the motion is at constant speed, then use P = F v with the balancing force.
A cyclist rides on level ground at constant speed $v$ against resistive force $F_r$ . Cyclist A rides at speed $v$ ; Cyclist B rides at speed $3v$ , with the same $F_r$ in each case. Which requires greater power?
Answer: Cyclist B, because $P=F_r v$ and tripling $v$ triples power. This question assesses the concept of power in AP Physics 1, which is the rate at which work is done or energy is transferred. At constant speed, power to overcome resistance is P = $F_r$ v, where $F_r$ is the resistive force. Both cyclists face the same $F_r$ , but Cyclist B at 3v requires P = $F_r$ (3v) = 3 $F_r$ v, triple that of Cyclist A's $F_r$ v. This assumes $F_r$ is independent of speed, as stated. Choice C misleads by claiming equal power from equal force, ignoring velocity's multiplication in the formula. To handle resistance problems, apply P = F v and note how changes in velocity affect power requirements.
A winch pulls a box up a ramp at constant speed. In trial 1, the tension does work $W$ in time $t$ . In trial 2, the tension does the same work $W$ in time $t/4$ . Which statement is correct?
Answer: Trial 2 has four times the average power because the same work is done in one-fourth the time. This question tests understanding of how power scales with time for constant work. Power is defined as P = W/t, showing an inverse relationship with time. In trial 1, P₁ = W/t, while in trial 2, P₂ = W/(t/4) = 4W/t = 4P₁. Trial 2 has four times the power because it completes the same work in one-fourth the time. Choice D incorrectly inverts the relationship, claiming smaller time means less power. When work is constant, power varies inversely with time: halving time doubles power, quartering time quadruples power.
A box is pulled at constant speed $v$ by a horizontal force $F$ for time $t$ . What is the average power delivered by $F$ ?
Answer: $Fv$ This question tests the concept of average power in the context of constant velocity motion. Power is defined as the rate at which work is done, given by P = W/t, where W is work and t is time. In this scenario, the work done by the force F over distance d = v t is W = F d = F v t, so average power P = (F v t)/t = F v. This shows that power depends on force and velocity, not directly on time for the average over that interval. A common distractor like choice A, Ft, might confuse power with impulse, which is force times time, but power involves energy transfer rate, not momentum change. To approach similar problems, always derive power from work divided by time or use P = F v for constant speed cases.
A student pushes a crate at constant speed $v$ across a rough floor with force $F$ for time $t$ . What is the average power delivered by the student?
Answer: $Fv$ This question tests understanding of power as the rate of energy transfer. Power is defined as work done per unit time, or P = W/t. Since the crate moves at constant speed, the applied force F equals the friction force, and the work done is W = F·d where d is the distance traveled. With constant speed v, the distance is d = vt, so W = Fvt. Therefore, power P = W/t = Fvt/t = Fv. Choice A (F/t) incorrectly treats force as work, while choice C (Ft) represents impulse, not power. When finding power with constant force and velocity, multiply force times velocity: P = Fv.
A battery transfers energy to a device at a constant rate. In $5\ \text{s}$ it transfers $E$ . In $10\ \text{s}$ it transfers $2E$ . What happens to average power?
Answer: It stays the same because $\frac{E}{5} = \frac{2E}{10}$ This question tests understanding of average power with constant energy transfer rate. Power is the rate of energy transfer, so average P = ΔE / Δt. In 5 s, P = E/5, and in 10 s, P = 2E/10 = E/5, remaining the same as expected for constant rate. This shows average power is consistent when energy scales linearly with time. Choice A is a distractor, mistakenly thinking more total energy means higher power without considering extended time. For such problems, check if rates are constant by seeing if energy over time is proportional.

Start this quiz